Linear equations with parameter. Linear equations with a parameter how to find the value of the parameter
TO tasks with parameter It can be attributed, for example, a search for solving linear and square equations in general form, the study of the equation to the number of roots available depending on the value of the parameter.
Do not bring detailed definitions, as examples, consider the following equations:
y \u003d kx, where x, y - variables, k - parameter;
y \u003d kx + b, where x, y are variables, k and b - parameter;
ax 2 + BX + C \u003d 0, where X is variables, A, B and C - parameter.
Solve equation (inequality, system) with a parameter This means, as a rule, to solve the infinite set of equations (inequality, systems).
The tasks with the parameter can be divided into two types:
but)the condition says: to solve the equation (inequality, system) is that it means for all the parameter values \u200b\u200bto find all solutions. If at least one case remained unexplored, it is impossible to recognize such a decision.
b)you need to specify the possible values \u200b\u200bof the parameter in which the equation (inequality, system) has certain properties. For example, it has one solution, does not have solutions, it has solutions belonging to the gap, etc. In such tasks, it is necessary to clearly indicate, with what the parameter value required by the desired condition is performed.
The parameter, being an unknown fixed number, has a special duality. First of all, it must be borne in mind that alleged fame says that the parameter must be perceived as a number. In second place, freedom of handling the parameter is limited to its unknown. For example, the fission operations on the expression in which the parameter or extraction of the root of an even degree is present from such an expression require preliminary studies. Therefore, accuracy in handling the parameter is necessary.
For example, to compare two numbers -6a and 3a, it is necessary to consider three cases:
1) -6a will be greater than 3a, if a negative number;
2) -6a \u003d 3a in the case when a \u003d 0;
3) -6a will be less than 3a, if a is the number positive 0.
The solution will be the answer.
Let the equation kx \u003d b be given. This equation is a brief record of an infinite set of equations with one variable.
In solving such equations, there may be cases:
1. Let K be any valid number of non-zero and B - any number of Isr, then x \u003d b / k.
2. Let k \u003d 0 and b ≠ 0, the initial equation will take a form 0 · x \u003d b. Obviously, there is no solution solutions.
3. Let k and b numbers equal to zero, then we have equality 0 · x \u003d 0. Its solution is any valid number.
Algorithm for solving this type of equations:
1. Determine the "control" parameter values.
2. Solve the initial equation with respect to x at the values \u200b\u200bof the parameter that were defined in the first paragraph.
3. Solve the initial equation with respect to x with the values \u200b\u200bof the parameter differ from the selected in the first paragraph.
4. Record the answer can be as follows:
1) at ... (parameter values), the equation has a root ...;
2) at ... (parameter values), no root equations.
Example 1.
Solve equation with parameter | 6 - x | \u003d a.
Decision.
It is easy to see that here a ≥ 0.
According to the rule of the module 6 - x \u003d ± a, we will express x:
Answer: x \u003d 6 ± A, where a ≥ 0.
Example 2.
Solve equation a (x - 1) + 2 (x - 1) \u003d 0 relative to the variable x.
Decision.
Recall brackets: Ah - a + 2x - 2 \u003d 0
We write the equation in standard form: x (a + 2) \u003d a + 2.
In case the expression A + 2 is not zero, i.e., if a ≠ -2, we have a solution x \u003d (a + 2) / (a \u200b\u200b+ 2), i.e. x \u003d 1.
In case a + 2 is zero, i.e. A \u003d -2, we have faithful equality 0 · x \u003d 0, so x - any valid number.
Answer: x \u003d 1 at a ≠ -2 and x € r at a \u003d -2.
Example 3.
Solve equation x / a + 1 \u003d a + x relative to the variable x.
Decision.
If a \u003d 0, we transform the equation to the form of a + x \u003d a 2 + ah or (a - 1) x \u003d -a (a - 1). The last equation at a \u003d 1 has the form 0 · x \u003d 0, therefore, x - any number.
If a ≠ 1, then the last equation will take the form x \u003d -a.
This solution can be illustrated on the coordinate direct (Fig. 1)
Answer: No solutions for a \u003d 0; x - any number at a \u003d 1; x \u003d -a at a ≠ 0 and a ≠ 1.
Graphic method
Consider another way to solve equations with a parameter - graphic. This method is used quite often.
Example 4.
How many roots depending on the parameter A has equation || x | - 2 | \u003d a?
Decision.
To solve a graphic method, we build graphs of functions Y \u003d || X | - 2 | and y \u003d a (Fig. 2).
Possible cases of direct y \u003d a and the number of roots in each of them are clearly visible in the drawing.
Answer: the roots of the equation will not be if< 0; два корня будет в случае, если a > 2 and a \u003d 0; Three roots equation will have in case a \u003d 2; Four root - at 0< a < 2.
Example 5.
With what equation 2 | x | + | x - 1 | \u003d A has the only root?
Decision.
Picture graphs of functions Y \u003d 2 | x | + | x - 1 | and y \u003d a. For y \u003d 2 | x | + | x - 1 |, open modules by the method of intervals, we get:
(-3x + 1, with x< 0,
y \u003d (x + 1, at 0 ≤ x ≤ 1,
(3X - 1, with x\u003e 1.
On the figure 3. It is clearly seen that the only root equation will only have at a \u003d 1.
Answer: A \u003d 1.
Example 6.
Determine the number of solutions of the equation | X + 1 | + | x + 2 | \u003d a depending on the parameter or?
Decision.
Schedule function y \u003d | x + 1 | + | x + 2 | will be a broken. Its vertices will be located at points (-2; 1) and (-1; 1) (Figure 4).
Answer: If the parameter A is less than one, the roots of the equation will not be; If a \u003d 1, then the solution of the equation is an infinite set of numbers from the segment [-2; -one]; If the values \u200b\u200bof the parameter A will be greater than one, the equation will have two roots.
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View equation f.(x.; a.) \u003d 0 called by the variable equation h. and parameter but.
Solve equation with parameter but - It means for each value but Find values h.satisfying this equation.
Example 1. oh= 0
Example 2. oh = but
Example 3.
x + 2 \u003d ah
x - ah \u003d -2
X (1 - a) \u003d -2
If 1 - but \u003d 0, i.e. but \u003d 1, then h.0 \u003d -2 roots no
If 1 - but 0, i.e. but 1, T. h. =
Example 4.
(but 2 – 1) h. = 2but 2 + but – 3
(but – 1)(but + 1)h. = 2(but – 1)(but – 1,5)
(but – 1)(but + 1)h. = (1but – 3)(but – 1)
If a but\u003d 1, then 0 h. = 0
h. - any valid number
If a but \u003d -1, then 0 h. = -2
No roots
If a but 1, but -1, T. h. \u003d (Single decision).
This means that every valid value but corresponds to the only value h..
For example:
if a but \u003d 5, then h. = = ;
if a but \u003d 0, then h. \u003d 3, etc.
Didactic material
1. oh = h. + 3
2. 4 + oh = 3h. – 1
3. but = +
for but \u003d 1 roots no.
for but \u003d 3 roots no.
for but = 1 h. - any valid number except h. = 1
for but = -1, but \u003d 0 solutions no.
for but = 0, but \u003d 2 solutions no.
for but = -3, but = 0, 5, but \u003d -2 solutions no
for but = -from, from \u003d 0 solutions no.
Square equations with parameter
Example 1. Solve equation
(but – 1)h. 2 = 2(2but + 1)h. + 4but + 3 = 0
For but = 1 6h. + 7 = 0
When but 1 highlight the values \u200b\u200bof the parameter in which D. Drawn to zero.
D \u003d (2 (2 but + 1)) 2 – 4(but – 1)(4but + 30 = 16but 2 + 16but + 4 – 4(4but 2 + 3but – 4but – 3) = 16but 2 + 16but + 4 – 16but 2 + 4but + 12 = 20but + 16
20but + 16 = 0
20but = -16
If a but < -4/5, то D. < 0, уравнение имеет действительный корень.
If a but \u003e -4/5 I. but 1, T. D. > 0,
h. = 
If a but \u003d 4/5, then D. = 0,
Example 2. At what values \u200b\u200bof the parameter and the equation
x 2 + 2 ( but + 1)h. + 9but - 5 \u003d 0 has 2 different negative root?
D \u003d 4 ( but + 1) 2 – 4(9but – 5) = 4but 2 – 28but + 24 = 4(but – 1)(but – 6)
4(but – 1)(but – 6) > 0
by t. Vieta: h. 1 + h. 2 = -2(but + 1)
h. 1 h. 2 = 9but – 5
By condition h. 1 < 0, h. 2 < 0 то –2(but + 1) < 0 и 9but – 5 > 0
| Eventually | 4(but – 1)(but – 6) > 0 - 2(but + 1) < 0 9but – 5 > 0 |
but < 1: а > 6 but > - 1 but > 5/9 |
 (Fig. one) < a. < 1, либо a. > 6 |
Example 3. Find values butunder which this equation has a solution.
x 2 - 2 ( but – 1)h. + 2but + 1 = 0
D \u003d 4 ( but – 1) 2 – 4(2but + 10 = 4but 2 – 8but + 4 – 8but – 4 = 4but 2 – 16but
4but 2 – 16 0
4but(but – 4) 0
but( but – 4)) 0
but( but – 4) = 0
a \u003d 0 or but – 4 = 0
but = 4
(Fig. 2.)
Answer: but 0 I. but 4
Didactic material
1. With what value but the equation oh 2 – (but + 1) h. + 2but - 1 \u003d 0 has one root?
2. With what value but the equation ( but + 2) h. 2 + 2(but + 2)h. + 2 \u003d 0 has one root?
3. At what values \u200b\u200band the equation ( but 2 – 6but + 8) h. 2 + (but 2 – 4) h. + (10 – 3but – but 2) \u003d 0 has more than two roots?
4. At what values \u200b\u200band equation 2 h. 2 + h. – but \u003d 0 has at least one common root with equation 2 h. 2 – 7h. + 6 = 0?
5. At what values \u200b\u200band the equations h. 2 +oh + 1 \u003d 0 and h. 2 + h. + but \u003d 0 have at least one common root?
1. Ply but = - 1/7, but = 0, but = 1
2. Ply but = 0
3. Ply but = 2
4. Ply but = 10
5. Ply but = - 2
Indicative equations with parameter
Example 1.. Sight all values butin which equation
9 x - ( but + 2) * 3 x-1 / x +2 but* 3 -2 / x \u003d 0 (1) There is exactly two roots.
Decision. Multiplying both parts of equation (1) by 3 2 / x, we obtain the equivalent equation
3 2 (x + 1 / x) - ( but + 2) * 3 x + 1 / x + 2 but = 0 (2)
Let 3 x + 1 / x \u003d w., then equation (2) will take a view w. 2 – (but + 2)w. + 2but \u003d 0, or
(w. – 2)(w. – but) \u003d 0, from where w. 1 =2, w. 2 = but.
If a w. \u003d 2, i.e. 3 x + 1 / x \u003d 2 h. + 1/h. \u003d log 3 2, or h. 2 – h.log 3 2 + 1 \u003d 0.
This equation has no valid roots, since it D. \u003d log 2 3 2 - 4< 0.
If a w. = but. 3 x + 1 / x \u003d but that h. + 1/h. \u003d log 3. but, or h. 2 – H.log 3 A + 1 \u003d 0. (3)
Equation (3) has exactly two roots if and only when
D \u003d log 2 3 2 - 4\u003e 0, or | Log 3 A | \u003e 2.
If log 3 a\u003e 2, then but \u003e 9, and if log 3 a< -2, то 0 < but < 1/9.
Answer: 0.< but < 1/9, but > 9.
Example 2.. Under what values \u200b\u200band equation 2 2x - ( but -3) 2 x - 3 but \u003d 0 has solutions?
In order for the given equation to have solutions, it is necessary and enough for the equation t. 2 – (a -3) t. – 3a. \u003d 0 had at least one positive root. We will find roots on the Vieta Theorem: h. 1 = -3, h. 2 = but = >
a is a positive number.
Answer: Ply but > 0
Didactic material
1. Find all the values \u200b\u200bof A, in which the equation
25 x - (2 but + 5) * 5 x-1 / x + 10 but * 5 -2 / x \u003d 0 has exactly 2 solutions.
2. Under what values \u200b\u200band the equation
2 (A-1) x? +2 (a + 3) x + a \u003d 1/4 has the only root?
3. At what values \u200b\u200bof the parameter and the equation
4 x - (5 but-3) 2 x +4 but 2 – 3but \u003d 0 has a single solution?
Logarithmic equations with parameter
Example 1. Find all values butin which equation
lOG 4X (1 + oh) = 1/2 (1)
it has a single decision.
Decision. Equation (1) is equivalent to equation
1 + oh = 2h. for h. > 0, h. 1/4 (3)
h. = w.
aU 2 - w. + 1 = 0 (4)
Not (2) condition from (3).
Let be but 0, T. aU 2. – 2w. + 1 \u003d 0 has valid roots if and only when D. = 4 – 4but 0, i.e. for but 1. To solve the inequality (3), we construct graphs of functions Galitsky M.L., Moshkovich M.M., Schwarzburg S.I.In-depth study of the course algebra and mathematical analysis. - M.: Enlightenment, 1990
The use of equations is widespread in our lives. They are used in many calculations, construction of structures and even sports. The person's equations used in antiquity and since then their application is only increasing. In mathematics, there are tasks in which it is necessary to search for solutions of linear and square equations in general form or search for the number of roots, which has an equation depending on the value of the parameter. All these tasks with parameters.
Consider the following equations as a visual example:
\\ [y \u003d kx, \\] where \\ - variables, \\ - parameter;
\\ [y \u003d kx + b, \\] where \\ - variables, \\ - parameter;
\\ [Ax ^ 2 + BX + C \u003d 0, \\] where \\ is the variable, \\ [A, B, C \\] - the parameter.
Solve equation with the parameter means, as a rule, to solve the infinite set of equations.
However, adhering to a certain algorithm, such equations can be easily solved:
1. Determine the "control" values \u200b\u200bof the parameter.
2. Solve the initial equation relative to [\\ x \\] with the values \u200b\u200bof the parameter defined in the first paragraph.
3. Solve the initial equation relative to [\\ x \\] with the values \u200b\u200bof the parameter differ from the selected in the first paragraph.
Suppose such an equation is given:
\\ [\\ MID 6 - X \\ MID \u003d a. \\]
After analyzing the source data, it can be seen that A \\ [\\ GE 0. \\]
According to the rule of the module \\ express \\
Answer: \\ where \\
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Under what values \u200b\u200bof the parameter $ a $ inequality $ () - x ^ 2 + (a + 2) x - 8a - 1\u003e 0 $ has at least one solution?
Decision
We present this inequality to the positive ratio with $ x ^ $ 2:
$ () - x ^ 2 + (a + 2) x - 8a - 1\u003e 0 \\ quad \\ leftrightarrow \\ quad x ^ 2 - (a + 2) x + 8a + 1< 0 .$
Calculate discriminant: $ d \u003d (a + 2) ^ 2 - 4 (8a + 1) \u003d a ^ 2 + 4a + 4 - 32a - 4 \u003d a ^ 2 - 28a $. So that this inequality has a solution, it is necessary that at least one point of parabola lay below the $ x $ axis. Since the branches of parabola are directed up, then for this you need to have a square three-root in the left side of the inequality, that is, its discriminant was positive. We come to the need to solve the square inequality $ a ^ 2 - 28a\u003e 0 $. The square three decreases $ a ^ 2 - 28a $ has two roots: $ a_1 \u003d 0 $, $ a_2 \u003d 28 $. Therefore, the inequality $ A ^ 2 - 28A\u003e 0 $ satisfy the gaps $ a \\ in (- \\ infty; 0) \\ Cup (28; + \\ infty) $.
Answer. $ a \\ in (- \\ infty; 0) \\ Cup (28; + \\ infty) $.
Under what values \u200b\u200bof the parameter $ a $ equation $ (a-2) x ^ 2-2ax + a + 3 \u003d $ 0 $ has at least one root, and at the same time all the roots are positive?
Decision
Let $ a \u003d $ 2. Then the equation takes a form $ () - 4x +5 \u003d 0 $, from where we get that $ x \u003d \\ dfrac (5) (4) $ is a positive root.
Now let $ A \\ Ne $ 2. The square equation is obtained. We define first at what values \u200b\u200bof the parameter $ a $ this equation has a root. It is necessary that its discriminant is non-negative. I.e:
$ D \u003d 4a ^ 2 - 4 (A-2) (a + 3) \u003d () -4a + 24 \\ geqslant 0 \\ leftrightarrow a \\ leqslant 6. $
The roots under the condition should be positive, therefore, from the Vieta Theorem we get the system:
$ \\ Begin (Cases) x_1 + x_2 \u003d \\ dfrac (2a) (a - 2)\u003e 0, \\\\ x_1x_2 \u003d \\ dfrac (a + 3) (a - 2)\u003e 0, \\\\ A \\ Leqslant 6 \\ End (Cases) \\ Quad \\ leftrightarrow \\ quad \\ begin (Cases) A \u200b\u200b\\ in (- \\ infty; 0) \\ Cup (2; + \\ infty), \\\\ A \\ in (- \\ infty; -3) \\ CUP ( 2; + \\ infty), \\\\ a \\ in (- \\ infty; 6] \\ END (Cases) \\ quad \\ leftrightarrow \\ quad a \\ in (- \\ infty; -3) \\ Cup (2; 6]. $
We combine the answers, we get the desired set: $ A \\ in (- \\ infty; -3) \\ Cup $.
Answer. $ A \\ in (- \\ infty; -3) \\ Cup $.
Under what values \u200b\u200bof the value of $ a $ inequality $ AX ^ 2 + 4AX + 5 \\ leqslant 0 $ does not have solutions?
Decision
- If $ a \u003d 0 $, then this inequality is degenerated into an inequality of $ 5 \\ leqslant 0 $, which has no solutions. Therefore, the value of $ a \u003d 0 $ satisfies the condition of the problem.
- If $ A\u003e 0 $, then the square is three-shredded in the left part of the inequality - parabola with branches directed upwards. Calculate $ \\ dfrac (d) (4) \u003d 4a ^ 2 - 5a $. Inequality does not have solutions if Parabola is located above the abscissa axis, that is, when the square triple has no roots ($ D< 0$). Решим неравенство $4a^2 - 5a < 0$. Корнями квадратного трёхчлена $4a^2 - 5a$ являются числа $a_1 = 0$ и $a_2 = \dfrac{5}{4}$, поэтому $D < 0$ при $0 < a < \dfrac{5}{4}$. Значит, из положительных значений $a$ подходят числа $a \in \left(0; \dfrac{5}{4}\right)$.
- If $ A.< 0$, то график квадратного трехчлена в левой части неравенства - парабола с ветвями, направленными вниз. Значит, обязательно найдутся значения $х$, для которых трёхчлен отрицателен. Следовательно, все значения $a < 0$ не подходят.
Answer. $ a \\ in \\ left $ lies between the roots, so the roots should be two (it means $ A \\ Ne $ 0). If the branches of parabolla is $ y \u003d ax ^ 2 + (a + 3) x - 3a $ are directed up, then $ y (-1)< 0$ и $y(1) < 0$; если же они направлены вниз, то $y(-1) > 0 $ and $ y (1)\u003e 0 $.
Case I. Let $ a\u003e 0 be $. Then
$ \\ left \\ (\\ begin (array) (L) y (-1) \u003d a- (a + 3) -3a \u003d -3a-3<0 \\ y(1)=a+(a+3)-3a=-a+3<0 \\ a>0 \\ END (Array) \\ Right. \\ Quad \\ leftrightarrow \\ quad \\ left \\ (\\ begin (array) (L) A\u003e -1 \\\\ A\u003e 3 \\\\ A\u003e 0 \\ END (Array) \\ Right. \\ Quad \\ Leftrightarrow \\ Quad A\u003e 3. $
That is, in this case, it turns out that all $ a\u003e $ 3 is suitable.
Clear II. Let $ A.< 0$. Тогда
$ \\ left \\ (\\ begin (array) (l) y (-1) \u003d a- (a + 3) -3a \u003d -3a-3\u003e 0 \\\\ y (1) \u003d a + (a + 3) -3a \u003d -a + 3\u003e 0 \\\\ A<0 \end{array} \right.\quad \Leftrightarrow \quad \left\{ \begin{array}{l} a<-1 \\ a<3 \\ a<0 \end{array} \right.\quad \Leftrightarrow \quad a<-1.$
That is, in this case, it turns out that all $ a< -1$.
Answer. $ a \\ in (- \\ infty; -1) \\ Cup (3; + \\ infty) $
Find all the values \u200b\u200bof the $ a $ parameter, each of which the equation system
$ \\ begin (Cases) x ^ 2 + y ^ 2 \u003d 2a, \\\\ 2XY \u003d 2A-1 \\ End (Cases) $
it has exactly two solutions.
Decision
Submount from the first second: $ (x-y) ^ 2 \u003d 1 $. Then
$ \\ left [\\ begin (array) (L) x-y \u003d 1, \\\\ x-y \u003d -1 \\ end (array) \\ Right. \\ quad \\ leftrightarrow \\ quad \\ left [\\ begin (array) (L) x \u003d y + 1, \\\\ x \u003d y-1. \\ End (Array) \\ Right. $
Substituting the obtained expressions into the second equation of the system, we get two square equations: $ 2y ^ 2 + 2y - 2a + 1 \u003d 0 $ and $ 2y ^ 2 - 2y - 2a + 1 \u003d 0 $. The discriminant of each of them is $ d \u003d 16a-4 $.
We note that it cannot happen that the pair of the first of the square equations coincides with the pair of the roots of the second square equation, since the amount of the roots of the first is $ -1 $, and the second 1.
So, it is necessary that each of these equations have one root, then the initial system has two solutions. That is, $ d \u003d 16a - 4 \u003d 0 $.
Answer. $ a \u003d \\ dfrac (1) (4) $
Find all the values \u200b\u200bof the $ a $ parameter, each of which equation $ 4x- | 3x- | x + a || \u003d 9 | X-3 | $ has two roots.
Decision
Refer to the equation in the form:
$ 9 | x-3 | -4x + | 3x- | x + a || \u003d 0. $
Consider the function $ f (x) \u003d 9 | x-3 | -4x + | 3x- | x + a || $.
With $ x \\ geqslant $ 3, the first module is revealed with a plus sign, and the function takes the form: $ F (x) \u003d 5x-27 + | 3x- | x + A || $. Obviously, with any disclosure of modules, a linear function with a coefficient of $ k \\ geqslant 5-3-1 \u003d 1\u003e 0 $ will be obtained, that is, this function at this gap is indefinitely increasing.
Consider now the gap of $ x<3$. В этом случае первый модуль раскрывается с минусом, и функция принимает следующий вид: $f(x) = - 13x+27+|3x-|x+a||$. При любом раскрытии модулей в итоге будет получаться линейная функция с коэффициентом $k\leqslant - 13+3+1 = - 9<0$, то есть на этом промежутке функция убывает.
So, we got that $ x \u003d 3 $ is the minimum point of this function. And this means that in order for the initial equation to have two solutions, the value of the function at the point of the minimum should be less than zero. That is, there is an inequality: $ F (3)<0$.
$ 12- | 9- | 3 + a ||\u003e 0 \\ quad \\ leftrightarrow \\ quad | 9- | 3 + a ||< 12 \quad \Leftrightarrow \quad -12 < 9-|3+a| < 12 \quad \Leftrightarrow \quad$
$ \\ Leftrightarrow \\ quad | 3 + a |< 21 \quad \Leftrightarrow \quad - 21 < 3+a < 21 \quad \Leftrightarrow \quad -24 Answer. $ a \\ in (-24; 18) $ Under what values \u200b\u200bof the parameter $ a $ equation $ 5 ^ (2x) -3 \\ CDot 5 ^ x + a-1 \u003d 0 $ is the only root? We will replace: $ T \u003d 5 ^ x\u003e 0 $. Then the initial equation takes the view of the square equation: $ t ^ 2-3t + a - 1 \u003d 0 $. The initial equation will have the only root in the event that this equation has one positive root or two roots, one of which is positive, the other is negative. The equation discriminant is: $ d \u003d 13-4a $. One root This equation will have in the event that the resulting discriminant turns out to be zero, that is, with $ a \u003d \\ dfrac (13) (4) $. In this case, the root $ t \u003d \\ dfrac (3) (2)\u003e 0 $, so this value of $ a $ is suitable. If there are two roots, one of which is positive, the other is non-positive, then $ d \u003d 13-4a\u003e 0 $, $ x_1 + x_2 \u003d 3\u003e 0 $ and $ x_1x_2 \u003d a - 1 \\ leqslant 0 $. That is, $ a \\ in (- \\ infty; 1] $ Answer. $ a \\ in (- \\ infty; 1] \\ CUP \\ LEFT \\ (\\ DFRAC (13) (4) \\ Right \\) $ Find all the values \u200b\u200bof the $ a $ parameter in which the system $ \\ begin (Cases) \\ log_a y \u003d (x ^ 2-2x) ^ 2, \\\\ x ^ 2 + y \u003d 2x \\ End (Cases) $ it has exactly two solutions. We transform the system to the following form: $ \\ begin (Cases) \\ log_a y \u003d (2x-x ^ 2) ^ 2, \\\\ y \u003d 2x-x ^ 2. \\ End (Cases) $ Since the $ a $ parameter is at the base of the logarithm, the following restrictions are superimposed on it: $ A\u003e 0 $, $ A \\ NE $ 1. Since the variable $ y $ is the argument of logarithm, then $ y\u003e 0 $. Combining both equations of the system, go to the equation: $ \\ log_a y \u003d y ^ $ 2. Depending on what values \u200b\u200bis received by the $ a $ parameter, two cases are possible: Answer. $ a \\ in (0; 1) $ Consider the case when $ a\u003e $ 1. Since with a large value of $ t $ the function of the function $ f (t) \u003d a ^ T $ lies above the direct $ G (T) \u003d T $, then the only common point can only be a touch point. Let $ T_0 be a touch point. At this point, the derivative to $ F (T) \u003d A ^ T $ is equal to the unit (tangent of tilt angle), in addition, the values \u200b\u200bof both functions coincide, that is, the system takes place: $ \\ Begin (Cases) A \u200b\u200b^ (T_0) \\ ln a \u003d 1, \\\\ A ^ (T_0) \u003d T_0 \\ END (Cases) \\ quad \\ leftrightarrow \\ quad \\ begin (Cases) A \u200b\u200b^ (T_0) \u003d \\ DFRAC (1) (\\ ln a), \\\\ a ^ (\\ Tau) \u003d \\ Tau \\ End (Cases) $ Where $ T_0 \u003d \\ DFRAC (1) (\\ ln a) $. $ A ^ (\\ FRAC (1) (\\ ln a)) \\ ln a \u003d 1 \\ quad \\ leftrightarrow \\ quad a ^ (\\ log_a e) \u003d \\ FRAC (1) (\\ ln a) \\ quad \\ leftrightarrow \\ quad A \u003d E ^ (\\ FRAC (1) (E)). $ In this case, other common points in a straight and indicative function is obviously no. Answer. $ a \\ in (0; 1] \\ CUP \\ LEFT \\ (E ^ (E ^ (- 1)) \\ Right \\) $Decision
Decision
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